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Math Help - series

  1. #1
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    series

    Prove that each term of the following sequence is a perfect square:
    49 , 4489 , 444889 , 44448889............................
    Last edited by geniusgarvil; September 15th 2012 at 08:16 PM. Reason: my mistake
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  2. #2
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    Re: series

    Your very first number is NOT a perfect square! You obviously mean 49, not 44. And then each succeeding number is gained by putting a new "4" and "8" into the center. Have you looked at the square root of each of those numbers? That should give you a hint.
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  3. #3
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    Re: series

    Hello, geniusgarvil!

    Prove that each term of the following sequence is a perfect square:

    . . 49,\,4489,\,444889,\,44448889\,\hdots

    \text{The }4^{th}\text{ term is: }\:a_4 \;=\;4444(10^4) + 888(10) + 9

    . . . . . . . . . . . . =\;4(1111)(10^4) + 8(111)(10) + 9

    . . . . . . . . =\;4\left(\frac{10^4-1}{9}\right)(10^4) + 8\left(\frac{10^3-1}{9}\right)(10) + 9



    \text{The }n^{th}\text{ term is: }\:a_n \;=\;4\left(\frac{10^n-1}{9}\right)10^n + 8\left(\frac{10^{n-1}-1}{9}\right)10 + 9


    \text{We have: }\:\tfrac{4}{9}(10^{2n}) - \tfrac{4}{9}(10^n) + \tfrac{8}{9}(10^n) - \tfrac{8}{9}(10) + 9

    . . . . . =\;\tfrac{4}{9}(10^{2n}) + \tfrac{4}{9}(10^n) + \tfrac{1}{9}

    . . . . . =\;\tfrac{1}{9}\big[4(10^{2n}) + 4(10^n) + 1\big]

    . . . . . =\;\tfrac{1}{9}\big[2(10^n) + 1\big]^2

    . . . . . =\;\left[\frac{2(10^n) + 1}{3}\right]^2 \;\;\hdots\;\;\text{ a square!}
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  4. #4
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    Re: series

    wow, that's really amazing. Thanx
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